这是有效的 ANSI C++ 代码吗?试图在编译时生成结构成员的偏移量 [英] Is this valid ANSI C++ code? Trying to generate offsets of structure members at compile-time

问题描述

可能的重复:
是否来自 的 'offsetof' 宏;调用未定义的行为?
取消引用空指针

    #define _OFFS_OF_MEMBER(p_type, p_member) (size_t)&(((p_type *)NULL)->p_member)

    struct a 
    {
             int a, b;
    };

    size_t l = _OFFS_OF_MEMBER(struct a, b);

我与一些其他用户进行了一些聊天/对话，其中一个人说这是取消引用并访问地址 NULL 附近的地址空间.我说:获取成员的地址不会访问、触摸或读取该成员的值.根据标准，它是完全安全的.

I had a little chat/conversation with some fellow users, and one of them said that this is dereferencing and accessing the address space near address NULL. I said: taking an address of a member will not access, touch, or read the value of that member. According to standard it is completely safe.

    struct a* p = NULL;
    size_t offset = &p->b; // this may NOT touch b, it is not dereferencing
    // p->b = 0; // now, we are dereferincing: acccess violation time!

这是否始终是计算偏移量的安全方法，还是编译器可以根据标准自由取消引用并弄乱地址 NULL 附近的内存?

Is this always a safe way to calculate offset, or are compilers free to dereference and mess up the memory near address NULL according to standards?

我知道有一种安全的方法可以计算标准提供的偏移量，但我很好奇您对此有何看法.都赞成我的解释:对这个问题投赞成票 :-)

I know there is a safe way to calculate offsets provided by the standard, but I am curious what you have to say about this. All in favor of my explenation: up-vote this question :-)

推荐答案

这里没有取消引用任何无效的内容.该宏所做的只是告诉编译器在地址 NULL 处的内存中存在 p_type 类型的结构.然后它获取p_member 的地址，它是这个虚构结构的成员.所以，不要在任何地方取消引用.

You're not dereferencing anything invalid here. All that macro does is tell the compiler that a structure of type p_type exists in memory at the address NULL. It then takes the address of p_member, which is a member of this fictitious structure. So, no dereferencing anywhere.

事实上，这正是offsetof 宏，定义在 stddef.h 中.

正如一些评论所说，这可能不适用于 C++ 和继承，我只将 offsetof 与 C 中的 POD 结构一起使用.

As some of the comments say, this may not work well with C++ and inheritance, I've only used offsetof with POD structures in C.

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